Differential of a Function and Finding Stationary Points

[breen155]

Hey I'm stuck on a problem, i have to find the differential of y=5x(2x-1)^3 and therefore find the x-coordinate of a stationary point on a graph,
I use the chain rule and get the differential to be 30x(2x-1)^2 and therefore the x-coordinate to be 0.5
however my textbook says the differential is 5(2x-1)^2(8x-1) and the x-coordinate of the stationary point is 1/8th I was wondering if someone could explain how this answer was obtained. Thanks

 

[CompuChip]

You used the chain rule, but you forgot about the product rule?
If you open up the last pair of brackets the first term is precisely what you got.

 

[breen155]

Ah thank you very much but i was always taught to do the chain rule in equations like the one above but the product rule worked :P again, thanks.

 

[HallsofIvy]

Ah thank you very much but i was always taught to do the chain rule in equations like the one above but the product rule worked :P again, thanks.

Sometimes you need both!

 

[CompuChip]

If you have a product of two (or more things) you always start with the product rule,
(f(x) g(x))' = f'(x) g(x) + f(x) g'(x)

This obviously requires you to calculate f'(x) and g'(x) and you may need the chain rule for either f'(x), or g'(x), or both. For example, for the derivative of
(3x + 6)^2 (2x - 1)^3
you will need the product rule once and the chain rule twice.

 

[breen155]

I understand now. Thanks for all the help guys :)